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Question

Find the indicated probability using the Central Limit Theorem. From 1975 through 2020, the mean annual gain of the Dow Jones Industrial Average was 652 . A random sample of 33 years is selected from this population. What is the probability that the mean gain for the sample was between 500 and 700 ? Assume sigma=1541.

Answer 1

Hi Dylan,

Central Limit Theorem states that we can assume normality for large n--some books call large n 30, some say 50, some say 75. For this question, let's assume large n is 30 and we can apply the Central Limit Theorem.

Now, we are still working with a sample mean, which means we need standard error. The formula for this:

SE=sigma/sqrt(n) where:

sigma=population standard deviation

n=sample size

Here:

sigma=1541

n=33

SE=1541/sqrt(33)

SE=268.26

Now, we can apply the classic formula in statistics z=(x-mu)/sigma, since we are assuming normality with the central limit theorem, but we must use standard error in place of standard deviation since we are working with a sample. So:

z=(x-mu)/SE

Since we are working with two Dow gains--500 and 700--I will use x_1, x₂, z₁, and z₂ in my calculations.

z₁=(x₁-mu)/SE

x₁=500

mu=652

SE=268.26

z₁=(500-652)/268.26

z₁= -0.57

From z-table, P(Z<z₁)=P(X<x₁)=0.2843

Keep both z₁ and that probability in mind. Now let's find z₂.

z₂=(x₂-mu)/SE

x₂=700

mu=652

SE=268.26

z₂=(700-652)/268.26

z₂= 0.18

From z-table:

P(Z<z₂)=P(X<x₂)=0.5714

Now, when we have an inequality Pa<x<Pb, that means we are going to subtract the probabilities, so:

P= 0.5714 - 0.2843

P= 0.2871

High standard error lowers this probability because it ended up in a denominator. I hope this helps.